Solve the equation 4 cos^2 (x) + 5 cos x + 1 = 0 for ( 0 ≤ x ≤ 2 pi )

Question

4cos²(x) – 5cos(x) + 1 = 0 for ( 0 ≤ x ≤ 2 ∏ )

Answer 1

let y = cosx

4y^2 -5y + 1 = 0

(4y-1)(y-1) = 0

y= 1, 1/4

cosx = 1 or 1/4

x= cos^-1(1 or 1/4)

x = 0 or about 75.52 degrees or 284.48 degrees

actually, you have an ambiguity in the problem where you have both signs + and - for the 5cosx term

if you use the +5cosx term

then it becomes cosx = -1 or -1/4

then x = 180, 104.48, or 255.52 degrees